Question

Heat transfer is required to raise the temperature of a 0.699 kg aluminum pot containing 2.77 kg of water from 32.5ºC to the boiling point and then boil away 0.695 kg of water. If the rate of heat transfer is 292 W, how long does this process take? Aluminum's specific heat is 0.215 kcal/kgºC; Water's specific heat is 1.00 kcal/kgºC; and heat of vaporization for water is 539 kcal/kg.

Answer 1

Before boiling away 0.695 kg of water, both the aluminium pot and the water reach the boiling temperature 100⁰C. The heat required for pot and water to reach the boiling point of water is
Q1 = m_p C_pT +m_w C_wT
Where the subscripts p and w corresponds to pot and water respectively. m is the mass and C is the specific heat capacity. T is the change in temperature
C_p = 0.215 kcal/kg⁰C
C_w = 1.00 kcal/kg⁰C
Q₁ = 0.699 kg x (0.215 kcal/kg⁰C) x (100⁰C - 32.5⁰C) +2.77 kg x (1.00 kcal/kg⁰C) x (100⁰C - 32.5⁰C)
Q₁ = 197.1192375 kcal.
Next, we have to find the heat required to boil away 0.695 kg of water.

Q₂ = m L = 0.695 kg x (539 kcal/kg) =  374.605 kcal.
Total heat transferred = Q = Q₁ +Q₂ = 571.72 kcal.
The heat transferred in joules = 571.72 x 4184 Joules
Q = 2392094.21 J
The rate of heat transfer is 292 W = 292 J/s
The time required for the process = t = Q /rate
t = 2392094.21 J / 292 J/s
t = 8192.10 s